1 tim x
2x^4-6x^3+x^2+6x-3=0
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NX
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
1
28 tháng 11 2016
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
QT
3
19 tháng 7 2018
A) x3-6x2+12x-8=0
<=>(x-2)3=0
<=>x-2=0
<=>x=2
B)4(x-3)2 -(2x-1)(2x+1)=13
<=>4(x2-6x+9)-4x2+1=13
<=>4x2-24x+36-4x2+1=13
<=>-24x+37=13
<=>24x=37-13
<=>24x=24
<=>x=1
C)25x2-6(x+1)2=0
<=>(5x-\(\sqrt{6}\left(x+1\right)\))(5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}\left(x+1\right)\)=0 hoặc 5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}x-\sqrt{6}\)=0 <=>5x+\(\sqrt{6}x+\sqrt{6}\)=0
<=>x(5-\(\sqrt{6}\))=\(\sqrt{6}\) <=>x(5+\(\sqrt{6}\))=\(-\sqrt{6}\)
<=>x=\(\frac{\sqrt{6}}{5-\sqrt{6}}\) <=>x=\(\frac{-\sqrt{6}}{5+\sqrt{6}}\)
19 tháng 7 2018
Rút gọn C=(4+2A+A^2).(4-A^2).(4-2a+a^2) GIẢI GIÚP MIK ĐI
25 tháng 8 2016
d) (x - 2)^2 = 1
= x = 2 + 1 = 3
c) (x^2 + 1). (x + 2011) = 0
25 tháng 8 2016
Tim x:
a) x^2 + 2x = 0
= \(x^2+2x=0\)
= \(x^2=0:2=0\)
b) (x - 3) + 2x^2 - 6x = 0
Rút gọn thừa số chung :
\(2x^2-5x-3=0\)
x = \(\frac{-1}{2}\)x = 3
=\(x^2=0\)
=> x = 0
PT
4
20 tháng 8 2016
a) = (3x +1)2 =0
3x+1 =0
x = -1/3
b) = (5x)2 -22 =0
(5x+2)(5x-2) = 0
5x+2 =0
x = -2/5
5x -2 =0
x= 2/5
xem đi rui lam tip
LM
20 tháng 8 2016
a) 9x2 + 6x + 1 = 0 => (3x)2 + 2 x 3x + 1 = 0 => (3x + 1)2 = 0 => 3x + 1 = 0 => x = \(\frac{-1}{3}\)
b) 25x2 = 4 => x2 = 4 : 25 => x2 = 0,16 => x = 0,4 hoặc x = -0,4
c) 8 - 125x3 = 0 => 125x3 = 8 => x3 = 8 : 125 => x3 = \(\frac{8}{125}\)=> x = \(\frac{2}{5}\)
NT
1
23 tháng 5 2022
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
b: =>|17x-5|=|17x+5|
=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5
=>34x=0
hay x=0
c: =>|3x+4|=|4x-18|
=>4x-18=3x+4 hoặc 4x-18=-3x-4
=>x=22 hoặc 7x=14
=>x=22 hoặc x=2
12 tháng 5 2022
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
NH
3
9 tháng 10 2021
x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0
⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0
⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0
⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0
⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0
⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3
LM
9 tháng 10 2021
tl
x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0
⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0
⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0
⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0
⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0
⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3
^HT^
2x4 - 6x3 + x2 + 6x - 3 = 0
=> 2x4 - 2x3 - 4x3 + 4x2 - 3x2 + 3x + 3x - 3 = 0
=> 2x3(x - 1) - 4x2(x - 1) - 3x(x - 1) + 3(x - 1) = 0
=> (x - 1)(2x3 - 4x2 - 3x + 3) = 0
=> (x - 1)(2x3 + 2x2 - 6x2 - 6x + 3x + 3) = 0
=> (x - 1)[2x2(x + 1) - 6x(x + 1) + 3(x + 1)] = 0
=> (x - 1)(x + 1)(2x2 - 6x + 3) = 0
\(\Rightarrow\left[{}\begin{matrix}x-1\\x+1\\2x^2-6x+3\end{matrix}\right.\) (2x2 - 6x + 3 vô nghiệm)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)